8Matrix groups

IA Groups

8.4 Rotations and reflections in R

2

and R

3

Lemma. SO(2) consists of all rotations of R

2

around 0.

Proof.

Let

A ∈ SO

(2). So

A

T

A

=

I

and

det A

= 1. Suppose

A

=

a b

c d

.

Then

A

−1

=

d −b

−c a

.

So

A

T

=

A

−1

implies

ad − bc

= 1,

c

=

−b

,

d

=

a

.

Combining these equations we obtain

a

2

+

c

2

= 1. Set

a

=

cos θ

=

d

, and

c = sin θ = −b. Then these satisfies all three equations. So

A =

cos θ − sin θ

sin θ cos θ

.

Note that

A

maps (1

,

0) to (

cos θ, sin θ

), and maps (0

,

1) = (

− sin θ, cos θ

), which

are rotations by θ counterclockwise. So A represents a rotation by θ.

Corollary.

Any matrix in O(2) is either a rotation around 0 or a reflection in a

line through 0.

Proof. If A ∈ SO(2), we’ve show that it is a rotation. Otherwise,

A =

1 0

0 −1

cos θ − sin θ

sin θ cos θ

=

cos θ − sin θ

− sin θ − cos θ

since O(2) =

SO

(2)

∪

1 0

0 −1

SO

(2). This has eigenvalues 1

, −

1. So it is a

reflection in the line of the eigenspace

E

1

. The line goes through

0

since the

eigenspace is a subspace which must include 0.

Lemma. Every matrix in SO(3) is a rotation around some axis.

Proof.

Let

A ∈ SO

(3). We know that

det A

= 1 and

A

is an isometry. The

eigenvalues

λ

must have

|λ|

= 1. They also multiply to

det A

= 1. Since we are

in

R

, complex eigenvalues come in complex conjugate pairs. If there are complex

eigenvalues

λ

and

¯

λ

, then

λ

¯

λ

=

|λ|

2

= 1. The third eigenvalue must be real and

has to be +1.

If all eigenvalues are real. Then eigenvalues are either 1 or

−

1, and must

multiply to 1. The possibilities are 1

,

1

,

1 and

−

1

, −

1

,

1, all of which contain an

eigenvalue of 1.

So pick an eigenvector for our eigenvalue 1 as the third basis vector. Then in

some orthonormal basis,

A =

a b 0

c d 0

0 0 1

Since the third column is the image of the third basis vector, and by orthogonality

the third row is 0, 0, 1. Now let

A

0

=

a b

c d

∈ GL

2

(R)

with

det A

0

= 1.

A

0

is still orthogonal, so

A

0

∈ SO

(2). Therefore

A

0

is a rotation

and

A =

cos θ − sin θ 0

sin θ cos θ 0

0 0 1

in some basis, and this is exactly the rotation through an axis.

Lemma.

Every matrix in O(3) is the product of at most three reflections in

planes through 0.

Note that a rotation is a product of two reflections. This lemma effectively

states that every matrix in O(3) is a reflection, a rotation or a product of a

reflection and a rotation.

Proof.

Recall O(3) =

SO

(3)

∪

1 0 0

0 1 0

0 0 −1

SO

(3). So if

A ∈ SO

(3), we know

that

A

=

cos θ − sin θ 0

sin θ cos θ 0

0 0 1

in some basis, which is a composite of two

reflections:

A =

1 0 0

0 −1 0

0 0 1

cos θ sin θ 0

sin θ − cos θ 0

0 0 1

,

Then if

A ∈

1 0 0

0 1 0

0 0 −1

SO

(3), then it is automatically a product of three

reflections.

In the last line we’ve shown that everything in O(3)

\ SO

(3) can be written as

a product of three reflections, but it is possible that they need only 1 reflection.

However, some matrices do genuinely need 3 reflections, e.g.

−1 0 0

0 −1 0

0 0 −1